Assignment#4 Due: 5pm on the date stated in the course outline. Hand in to the assignment box on the 3 rd floor of CAB.

Transcription

1 MATH Assgmet#4 Due: 5pm o the date stated the course outle. Had to the assgmet box o the 3 rd floor of CAB.. Let deote the umber of teror regos of a covex polygo wth sdes, dvded by all ts dagoals, f o three dagoals pass through a commo pot sde the polygo. Use the theorem below to show that: for ( ) ( ) Theorem: Cosder a covex rego of the plae whch s crossed by les wth teror pots of tersecto. No three of the les pass through a commo pot sde the rego. The umber of dsjot regos created s: Soluto: There s oe teror pot of tersecto for every choce of four pots o the covex polygo, therefore ( ). There s oe le for every two pots o the covex polygo but of these les are ot dagoals, therefore ( ) Usg the theorem we have: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

2 . Use mathematcal ducto to prove that the sum of the frst square umbers s equal to That s, prove: Soluto: Base case: : ad, so the base case holds. Show: By ducto assumpto

3 3. Cojecture a formula for the sum of the frst Fboacc umbers wth odd dces ad prove your formula wors by usg mathematcal ducto. That s, fd ad prove a formula for Note: Soluto: Recall, F, F, F, F 3 4, F 5, F 8, F 3, F, F 34 9, F 55 3 The frst few terms gves us: 5 : : 3 3: : 5 3 5: Perhaps the sum of the frst Fboacc umbers wth odd dces s F. Base case: F F F, so the base case holds Iductve Step: Show: F F3 F5 F F F F F F F F 3 5 F F F F F F F F F 3 5 By ducto assumpto By Fboacc recurrece

4 4. A equal umber of ope ad closed gas statos are dstrbuted a equal dstace apart aroud a rg road of a cty. Say statos are ope ad statos are closed. Now suppose t taes gallos of gas to travel aroud the cty. If we start wth a empty ta ad ca oly add gallo at a tme at ay ope stato, show that t possble to fd a startg posto to travel aroud the cty clocwse for all? Soluto: Base case: Whe the s oly oe ope ad oe closed gas stato we ca start at the ope gas stato: O Iductve Step: C Show: f t possble to fd a startg posto to travel aroud the cty wth ope ad closed gas statos the t possble to fd a startg posto to travel aroud the cty wth ope ad closed gas statos. Cosder the rg road wth ope ad closed gas statos C O Search for a ope gas stato whch s followed by a closed gas stato. Notce that durg travel aroud the cty f oe adds a gallo at oe ca always travel oe gas stato past Notg that reduces the problem to fdg a startg posto amog ope ad closed gas statos: ad by our assumpto there s a startg posto. C O

5 5. Prove that a square ca be dssected to smaller squares for ay postve teger (The smaller squares do ot all have to be the same sze) For example a square ca be dssected to smaller squares: Soluto: Let be the statemet for Base Cases: The example gve shows s true. The followg two fgures are dssectos to 7 ad 8 squares respectvely: Therefore ad are true. Iductve Step: Show:. Notce that f a square ca be dssected to smaller squares, the t ca be dssected to smaller squares. Ths s doe by tag oe of the exstg squares ad dssectg t to four squares of equal sze. For example, usg the dssecto of squares, the followg s a dssecto to 9 squares: Therefore f s true the s also true.

6 . Let be the umber of words of legth wth dgts from the set wth o cosecutve s. a) Show that:,, s a recurrece relato for Soluto: Whe the words ad do ot have cosecutve s, therefore. For there are 8 words that do ot have cosecutve s:,,,,,,, therefore. To fd a recurrece relato for we loo at cases. CASE : The frst dgt s ot a zero. The the frst dgt s a or a ether case there are ways to complete the word wth o cosecutve s. a words a words CASE : The frst dgt s a zero. The the secod dgt must be a or ether case there are ways to complete the word wth o cosecutve s. a words a words Therefore:, s a recurrece relato for

7 b) Use strog ducto to show that: ( ) ( ) s a soluto to the recurrece relato from part a). Soluto: Let be the statemet for Base Cases: ( ) ( ) Therefore ad are true. ( ) ( ) Iductve Step: Show: &. ( ) ( ) ( ) ( ) (by P & P ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ) ( ( ) ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Therefore s true.

8 7. Prove Cass s Idetty: For Note: Soluto: Let be the statemet for Base Case: Therefore s true. Iductve Step: Show: ( by P ) Therefore s true.

9 Aother Soluto: For readers who have a lear algebra bacgroud. Frst otce that: * + [ ] * + [ ] * + [ ] * + [ ] Now tae the determat of both sdes: 8. The objectve of ths questo s to prove the followg theorem. For all atural umbers : a defcet board ca be tled wth rght tromoes. Recall: whe every defcet board ca be tled wth tromoes. Fsh provg the theorem by completg parts a) ad b). a) Show that f the a defcet board caot be tled wth tromoes. b) Show that f ad the a defcet board ca be tled wth tromoes.

10 Soluto: a) Note: a defcet board has a area of:. I order to tle ay board wth tromoes ts area must be dvsble by 3 but f the: Therefore, f a defcet board caot be tled wth tromoes. b) Note that the specal cases of have both bee show example of lecture, so we ca safely assume that Let for ad secto off a defcet board as follows: A B C D Now by symmetry we oly eed to cosder the case whe the mssg square s secto. Each secto ca be tled wth tromoes for the followg reasos: Therefore, f ad a defcet board ca be tled wth tromoes.

11 9. For all show that ay defcet board ca be tled wth rght tromoes. Soluto: Base case: I ths case we have a board ad start by breag t to the followg sectos: 7 A B By symmetry we oly eed to cosder the case whe the mssg square s secto or CASE : the mssg square s secto A C Now:

12 CASE : the mssg square s secto Place a bloc aroud secto as follows: D E F Now, Iductve step: Show: f ay defcet board ca be tled wth tromoes the ay defcet board ca be tled wth tromoes (for. I a defcet the mssg square s the top left: board by symmetry we oly eed to cosder the case whe By placg a tromo the ceter as show we create four defcet whch ca be tled by ducto. boards

13 0. The prcple of double ducto s stated as follows. If s a statemet about the tegers ad such that. s true,. For all, f s true, the s true, 3. For all, f s true for all, the s true for all, the s true for all ad. Use the prcple of double ducto to show that for ay tegers ad we have: Soluto: Let be the statemet that for Step : Show: s true. Step : Show for Therefore s true. Step 3: Show for Therefore s true.

14 Bous. A game s played wth a dec of cards umbered from to. They are shuffled thoroughly ad the top card s tured over. If t s umber, the game s wo. If t s umber where the t s serted to the dec so that t s the -th card from the top. The the ew top card s tured over ad the same process s appled. Ca ths game be wo evetually, regardless of how the cards are staced? Soluto. The game ca always be wo by ducto. Base case: The card o the top s labeled ad the game s wo. Iductve Step: Show: the game s evetually wo wth a dec of sze the game s evetually wo wth a dec of sze Case : The card labeled comes to the top. After the card labeled comes to the top t gets placed o the bottom of the dec. The card labeled wll stay o the bottom for the rest of the game. I ths case the game s played wth the remag cards from the top ad s evetually wo by. Top Card cards labeled d Last card Does ot move

15 Case : The card labeled does ot come to the top. a) Suppose that the card labeled s o the bottom of the dec. Sce the card labeled does ot come to the top throughout the game o card wll be placed o the bottom of the dec throughout the game. Ths meas the card labeled wll stay o the bottom for the rest of the game. I ths case the game s played wth the remag cards from the top: Sppg Sce the card labeled does ot come to the top t does ot chage the way the game s played. Therefore we ca let the card labeled play the role of the card labeled ad play the game wth a dec of cards umbered from to. Now, the game s wo by. Top Card B cards labeled d Last Card B Does ot move Sppg B b) Suppose that the card labeled s o the bottom of the dec. Follow the argumet from part a). I dog so, we would let the card labeled play the role of the card labeled ad tur the card labeled comes to the top. But, our assumpto for ths case s that the card labeled does ot come to the top. Therefore Case b) results a cotradcto ad does ot occur.

16 Aother Soluto. Ths game has 3 types of players: Goal Scorers are cards that come to the top of the dec. Defeders are cards that do ot come to the top of the dec. Defeders move about the dec oly tag up space; as a result they ca chage ther ame. Bech Warmers are cards that are stuc o the bottom of the dec. Bech Warmers do ot affect the outcome so they ca be removed from the game whle they st below the cards stll the game. Note: Bech Warmers are also Defeders except the specal case of a dec of sze. I ths specal case card oe s both a Goal Scorer ad a Bech Warmer. Step Show: as the game s played every dec wll produce a Bech Warmer uless the game s wo beforehad. After a Bech Warmer s produced ths game wll be played wth less card. There are cases: Case. The largest card s a Goal Scorer. I ths case the largest card wll come to the top ad become a Bech Warmer. The ths game ca be played wth less card. Case. The largest card s a Defeder. I ths case there s a card stuc o the bottom of the dec; ths card s a Bech Warmer. Now, the largest card ad ths Bech Warmer swap ames ad the our game s played wth less card. Step Show: card umber oe s a Goal Scorer. Assume that card umber oe s a Defeder. The the game wll cotue beg played forever; but as the game s played every dec wll produce a Bech Warmer. After a Bech Warmer s produced ths game wll be played wth less card. Ths process (from step oe) wll cotue o utl all the cards are Bech Warmers. Whe all the cards are Bech Warmers the card o the top s both a Goal Scorer ad a Bech Warmer. Card oe s the oly card capable of such a feat. Ths cotradcts card oe beg a Defeder.